Question: $ F(x) = \int_{0}^{\sqrt{x}} 2t \, dt$ where $x>0$. $F'(x) = $
Solution: The following hints will explain how to solve this problem, but it's really important that we understand why this works. To get that understanding, we highly recommend that you watch Sal solve a similar problem. Here are the steps to solve this problem: $\phantom{=}F'(x)$ $ = \dfrac{d}{dx} \int_{0}^{\sqrt{x}} 2t \, dt$ $= 2\sqrt{x} \cdot \dfrac{d}{dx} \sqrt{x}$ $= 2\sqrt{x} \cdot \dfrac1{2\sqrt{x}}$ $= 1$